Tagged Questions

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How to find the missing number?

A teacher intended to give a typist a list of nine integers that form a group under multiplication modulo 91. But one of the nine integers was inadvertently left out, so that the list appeared as ...
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Prove that this equation have an non finite number of prime solutions

So the question seeks to answer the following, let $x,y\in\Bbb R$. Prove that there is a non finite number of prime solutions to the following equation: $3x-5y=11$. Our professor says that it's easy ...
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The relationship between the mod value of 2 numbers?

I've been trying to study on my own, the relationship between 2 numbers as you move down the number line from a starting point and could use some help. I believe this could be described as modular ...
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Finding the totient functionlike function for an irrational number like (a+b*sqrt(5)) where a and b are whole numbers mod M where M is a whole number.

I need to find if a value $T$ exists for irrational number of the form $(a+b\cdot \sqrt{5})$ such that $(a+b\cdot \sqrt{5})^T = 1 \pmod M$. Also ,is it possible to find out upper bound for T .
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What will be the multiplicative inverse of square root of 5 with respect to a natural number $M$?

Can such a number $N$ be found such that $\sqrt{5}N \equiv 1 \mod M$? If no,what can be the best approximation for $N$?
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How to Solve an equation with mod for a variable?

I have following equation to be solved, but I am having some trouble in making an understanding and doing so. (d * e) % v = 1 e and v are known. How to solve this ...
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difference between angles

i could not understand exactly what is asked in the following question: What is difference in the degree measures of the angles formed by Hour hand and minute Hand of a clock at $12:35$ and ...
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Finding the last two digits of the expansion of $2^{12n}-6^{4n}$

The question is: Find the last two digits of the expansion of $2^{12n}-6^{4n}$ where $n$ is any positive integer. If we put the value of $n=1$ we would get $2800$. For $n = 2$ the result will ...
Prove that $(n-m) \mid (n^r - m^r)$
In respect to a larger proof I need to prove that $(n-m) \mid (n^r - m^r)$ (where $\mid$ means divides, i.e., $a \mid b$ means that $b$ modulus $a$ = $0$). I have played around with this for a while ...
How do I find the lowest $n$ for which $a^n \equiv 1 \pmod{b}$?
This is mostly related to doing large modular exponentiation by hand. For example, a problem I was doing was to find the last 3 digits of $7^{9729}$; that is, find $7^{9729}\bmod{1000}$. Using the ...